Minimizing 1/2-harmonic maps into spheres

Abstract

In this article, we improve the partial regularity theory for minimizing 1/2-harmonic maps of Millot and Sire (Arch Ration Mech Anal 215:125–210, 2015), Moser( J Geom Anal 21:588–598, 2011) in the case where the target manifold is the $(m-1)$-dimensional sphere. For $m\geqslant 3$, we show that minimizing 1/2-harmonic maps are smooth in dimension 2, and have a singular set of codimension at least 3 in higher dimensions. For $m=2$, we prove that, up to an orthogonal transformation, x/|x| is the unique non trivial 0-homogeneous minimizing 1/2-harmonic map from the plane into the circle ${\mathbb {S}}^1$. As a corollary, each point singularity of a minimizing 1/2-harmonic map from a 2d domain into ${\mathbb {S}}^1$ has a topological charge equal to $\pm 1$.

Appendix A

We provide in this appendix some details about the computations performed in Sect. 5.3.

Lemma A.1

For every $\varvec{\gamma }\in [0,1)$,

$$\begin{aligned} I(\varvec{\gamma }):=\int _{{\mathbb {D}}} \frac{(1+|z|^2)^2-4z^2_1}{(1-2\varvec{\gamma }z_1+\varvec{\gamma }^2|z|^2)(1+|z|^2)^2} \,\mathrm{d}z=\pi F(\varvec{\gamma }^2) \end{aligned}$$

with

$$\begin{aligned} F(t):=\left( \frac{t^2-10t+1}{(1+t)^4}\right) \log \left( \frac{(1-t)^2}{4}\right) -\frac{t^2+11t-2}{(1+t)^3} \,. \end{aligned}$$

Proof

Write $I(\varvec{\gamma })=A(\varvec{\gamma }) -4B(\varvec{\gamma })$ with

$$\begin{aligned} A(\varvec{\gamma }):= \int _{{\mathbb {D}}} \frac{1}{(1-2\varvec{\gamma }z_1+\varvec{\gamma }^2|z|^2)}\,\mathrm{d}z \end{aligned}$$

and

$$\begin{aligned} B(\varvec{\gamma }):=\int _{{\mathbb {D}}} \frac{z^2_1}{(1-2\varvec{\gamma }z_1+\varvec{\gamma }^2|z|^2) (1+|z|^2)^2}\,\mathrm{d}z\,. \end{aligned}$$

Using polar coordinates, we further rewrite

$$\begin{aligned} A(\varvec{\gamma })=\int _0^1 M(\varvec{\gamma }r)r \,\mathrm{d}r\quad \text {and}\quad B(\varvec{\gamma })=\int _0^1 \frac{N(\varvec{\gamma }r)r^3}{(1+r^2)^2}\,\mathrm{d}r\,, \end{aligned}$$

where

$$\begin{aligned} M(a):= \int _0^{2\pi } \frac{\mathrm{d}\theta }{(1-2a\cos (\theta )+a^2)^2} \quad \text {and}\quad N(a):=\int _0^{2\pi }\frac{\cos ^2(\theta )}{(1-2a\cos (\theta )+a^2)^2}\,\mathrm{d}\theta \end{aligned}$$

are defined for $a\in [0,1)$.

Lengthy but elementary computations yield

$$\begin{aligned} M(a)=2\pi \frac{1+a^2}{(1-a^2)^3} \quad \text {and} \quad N(a)=2\pi \left( \frac{1+a^2}{(1-a^2)^3}-\frac{1}{2(1-a^2)}\right) \,. \end{aligned}$$

Then we first obtain

$$\begin{aligned} A(\varvec{\gamma })=2\pi \int _0^1\frac{\varvec{\gamma }^2r^3+r}{(1-\varvec{\gamma }^2r^2)^3}\,\mathrm{d}r=\pi \left[ \frac{r^2}{(1-\varvec{\gamma }^2r^2)^2}\right] ^1_0 =\frac{\pi }{(1-\varvec{\gamma }^2)^2}\,. \end{aligned}$$

(A.1)

Concerning $B(\varvec{\gamma })$, we can rewrite it as

$$\begin{aligned} B(\varvec{\gamma })=\pi \big (2U(\varvec{\gamma }^2)-V(\varvec{\gamma }^2)\big ) \end{aligned}$$

(A.2)

with

$$\begin{aligned} U(t):=\int _0^1\frac{(1+tr^2)r^3}{(1-tr^2)^3(1+r^2)^2} \,\mathrm{d}r\quad \text {and}\quad V(t):=\int _0^1\frac{r^3}{(1-tr^2)(1+r^2)^2}\,\mathrm{d}r\,. \end{aligned}$$

Once again, elementary computations lead to

$$\begin{aligned} V(t)=\frac{1}{2(1+t)^2}\log \left( \frac{2}{1-t}\right) -\frac{1}{4(1+t)} \end{aligned}$$

and

$$\begin{aligned} U(t)=\left( \frac{t^2-4t+1}{2(1+t)^4}\right) \log \left( \frac{2}{1-t}\right) +\frac{1}{8(1-t)^2}+\frac{1}{2}P(t)\,, \end{aligned}$$

with

$$\begin{aligned} P(t)&:=\frac{1}{4(1-t)} +\frac{1}{4(1+t)} -\frac{3}{4(1+t)^2} +\frac{t^2+2t}{(1+t)^2(1-t)}\\&\quad -\frac{t}{(1+t)(1-t)}-\frac{4t^2}{(1+t)^3(1-t)}-\frac{1-t}{2(1+t)^3}\,. \end{aligned}$$

Therefore,

$$\begin{aligned} 2U(t)-V(t)= \left( \frac{t^2-10t+1}{2(1+t)^4}\right) \log \left( \frac{2}{1-t}\right) +\frac{1}{4(1-t)^2}+P(t)+\frac{1}{4(1+t)}\,. \end{aligned}$$

(A.3)

A direct computation shows that

$$\begin{aligned} P(t)+\frac{1}{4(1+t)}=\frac{t^2+11t-2}{4(1+t)^3}\,. \end{aligned}$$

(A.4)

Gathering (A.1)–(A.2)–(A.3)–(A.4) now leads to $I(\varvec{\gamma })=\pi F(\varvec{\gamma }^2)$ as announced. $\square$

Lemma A.2

Let ${\mathfrak {C}}$ be the Cayley transform (defined in (5.38)). For $a\in (0,1]$ and $z\in \overline{{\mathbb {D}}}$, let

$$\begin{aligned} K_a(z):= \frac{|{\mathfrak {C}}^{-1}(z)|^2}{(a^2+2a{\mathfrak {C}}_2^{-1}(z) +|{\mathfrak {C}}^{-1}(z)|^2)^2}\,, \end{aligned}$$

where ${\mathfrak {C}}^{-1}_2$ denotes the imaginary part of ${\mathfrak {C}}^{-1}$. Define for $\lambda \in (0,1)$,

$$\begin{aligned} J(a,\lambda ):=\frac{1}{2\pi }\int _{{\mathbb {S}}^1} K_a(\lambda \sigma )\,\mathrm{d}\sigma \,. \end{aligned}$$

Then,

$$\begin{aligned} J(a,\lambda )= \frac{(1+t)^4}{16}\left( \frac{(2t^2+1)t^2 \lambda ^6-(6t^2-1)\lambda ^4+t^2\lambda ^2+1}{(1-\lambda ^2 t^2)^3}\right) \text { with }t:=\frac{1-a}{1+a} \end{aligned}$$

(A.5)

for every $\lambda \in [0,1]$. In addition, $\lambda \in [0,1]\mapsto J(a,\lambda )$ is increasing for every $a\in (0,1]$.

Proof

Recalling that

we change variables to obtain

$$\begin{aligned} \frac{1}{2\pi }\int _{{\mathbb {S}}^1}K_a(\lambda z) \,\mathrm{d}{\mathcal {H}}^1=\frac{1}{\pi }\int _{\mathbb {R}}\frac{K_a\big (\lambda {\mathfrak {C}}(x)\big )}{1+x^2}\,\mathrm{d}x\,. \end{aligned}$$

Next we set

$$\begin{aligned} c:=\frac{1-\lambda }{1+\lambda }\in (0,1)\,,\quad A:=\frac{a+c}{1+ac} \,,\quad B:=c^2+\frac{1}{c^2}\,, \end{aligned}$$

to compute

$$\begin{aligned} K_a\big (\lambda {\mathfrak {C}}(x)\big )=\left( \frac{c^2}{(1+ac)^4}\right) \frac{x^4+B x^2+1}{(x^2+A^2)^2}\,. \end{aligned}$$

By Lemma A.4 below, we have

$$\begin{aligned} J(a,\lambda )= \frac{c^2}{(1+ac)^4} \left( \frac{1+A^2}{2A^3} +\frac{B-2}{2A(A+1)^2}\right) \,. \end{aligned}$$

In terms of the variables t and $\mu :=\lambda ^2\in (0,1)$, we obtain

$$\begin{aligned} J(a,\lambda )=\frac{(1+t)^4}{16}\left( \frac{(2t^2+1)t^2\mu ^3-(6t^2-1) \mu ^2+t^2\mu +1}{(1-\mu t^2)^3}\right) \,, \end{aligned}$$

which is the announced formula. Next, if

$$\begin{aligned} f:\mu \in (0,1)\mapsto \frac{(2t^2+1)t^2\mu ^3 -(6t^2-1)\mu ^2+t^2\mu +1}{(1-\mu t^2)^3}\,, \end{aligned}$$

we have

$$\begin{aligned} f^\prime (\mu )= \frac{4t^2(1-\mu )^2+2\mu (1-t^2)^2}{(1-\mu t^2)^4}>0\,, \end{aligned}$$

which shows that $\lambda \mapsto J(a,\lambda )$ is indeed increasing for every $a\in (0,1)$. $\square$

Remark A.3

Note that the function $J(a,\lambda )$ defined in (A.5) can be rewritten as

$$\begin{aligned} J(a,\lambda )=\frac{(1+t)^4}{32}\left( \frac{(1-\lambda ^2)^2}{(1+\lambda t)(1-\lambda t)^3}+\frac{(1-\lambda ^2)^2}{(1+\lambda t)^3(1-\lambda t)} +\frac{4\lambda ^2}{1-\lambda ^2t^2}\right) \,. \end{aligned}$$

From this formula, one easily determines the behavior of J as $a\sim 0$ and $\lambda \sim 1$.

Lemma A.4

For $A,B>0$, we have

$$\begin{aligned} \frac{1}{\pi }\int _{\mathbb {R}}\frac{x^4+Bx^2+1}{(1+x^2)(x^2+A^2)^2} \,\mathrm{d}x=\frac{1+A^2}{2A^3}+\frac{B-2}{2A(A+1)^2}\,. \end{aligned}$$

(A.6)

Proof

Write $X:=x^2$, and observe that

$$\begin{aligned} \frac{X^2+BX+1}{(X+1)(X+A^2)^2}&=\frac{2-B}{(1-A^2)^2}\frac{1}{X+1}\\&\quad + \left( 1+\frac{B-2}{(1-A^2)^2}\right) \frac{1}{X+A^2}\\&\quad +\left( (1-A^2)+(2-B)\frac{A^2}{1-A^2}\right) \frac{1}{(X+A^2)^2} \,. \end{aligned}$$

On the other hand,

$$\begin{aligned} \frac{1}{\pi }\int _{\mathbb {R}}\frac{\mathrm{d}x}{1+x^2}=1\,,\quad \frac{1}{\pi } \int _{\mathbb {R}}\frac{\mathrm{d}x}{x^2+A^2}=\frac{1}{A}\,,\quad \frac{1}{\pi } \int _{\mathbb {R}}\frac{\mathrm{d}x}{(x^2+A^2)^2}=\frac{1}{2A^3}\,, \end{aligned}$$

and (A.6) follows. $\square$